proof of Bernstein inequalities
1) By Chernoff-Cramèr bound (http://planetmath.org/ChernoffCramerBound), we have:
where
Since ,
and, keeping in mind hypotheses a) and b),
Now, if , we obtain
whence
By elementary calculus, we obtain the value of that maximizes theexpression in brackets (out of the two roots of the second degree polynomial equation, we choose the one which is ):
which, once plugged into the bounds, yields
Observing that , one gets:
Plugging in the bound expression, the sub-optimal yet moreeasily manageable formula is obtained:
which is obviously a worse bound than the preceeding one, since . One can also verify the consistency of this inequalitydirectly proving that, for any ,
(see here (http://planetmath.org/ASimpleMethodForComparingRealFunctions) for an easy way, which can be used with )
2) To prove this more specialized statement let’s recall that the condition
implies that, for all ,
(See here (http://planetmath.org/RelationBetweenAlmostSurelyAbsolutelyBoundedRandomVariablesAndTheirAbsoluteMoments) for a proof.)
Now, it’s enough to verify that the condition
imply both conditions a) and b) in part 1).
Indeed, part a) is obvious, while for part b) one happens to have:
(see here (http://planetmath.org/AbsoluteMomentsBoundingNecessaryAndSufficientCondition) for a proof).
So
Let’s find a value for such that ,thus satisfying part b) of the hypotheses.
After simplifying, we have to study the inequality
for any . Let’s proceed by induction. For , we have
which suggests . Let’s now verify if this position isconsistent with the inductive hypothesis:
which confirms the validity of the choice , which has to beplugged into the former bound to obtain the new one.
[to be continued…]